1992 AMC 12 第 15 题

Question

令

i=\sqrt{-1}

。对

n\ge1

，按下式定义一个复数数列：

\begin{aligned} z_1&=0,\\ z_{n+1}&=z_n^2+i \end{aligned}\text{。}

在复平面中，

z_{111}

到原点的距离是多少？

Let

i=\sqrt{-1}.

For

n\ge1,

define a sequence of complex numbers by

\begin{aligned} z_1&=0,\\ z_{n+1}&=z_n^2+i. \end{aligned}

In the complex plane, how far from the origin is

z_{111}?

1

\sqrt2

\sqrt3

\sqrt{110}

\sqrt{2^{55}}

先计算数列的前几项，不要试图直接展开

z_{111}

Compute the first few terms rather than trying to expand

z_{111}

从

z_3

开始寻找循环

Look for a cycle beginning with

z_3

前几项为

\begin{aligned} z_1&=0,\quad z_2=i,\quad z_3=-1+i,\\ z_4&=-i,\quad z_5=-1+i \end{aligned}\text{。}

因此从

z_3

起，奇数下标项为

-1+i

，偶数下标项为

-i

，两者交替出现。所以

z_{111}=-1+i

，它到原点的距离为

\sqrt{(-1)^2+1^2}=\sqrt2

。

因此正确答案是 B。

The first terms are

\begin{aligned} z_1&=0,\quad z_2=i,\quad z_3=-1+i,\\ z_4&=-i,\quad z_5=-1+i. \end{aligned}

Hence the sequence alternates between

-1+i

at odd indices and

-i

at even indices from

z_3

onward. Therefore

z_{111}=-1+i,

whose distance from the origin is

\sqrt{(-1)^2+1^2}=\sqrt2.

Thus the correct answer is B.

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